ESTIMATES FOR THE GREEN FUNCTION AND
SINGULAR SOLUTIONS FOR POLYHARMONIC
NONLINEAR EQUATION
IMED BACHAR, HABIB MÂAGLI, SYRINE MASMOUDI,
AND MALEK ZRIBI
Received 5 July 2002
We establish a new form of the 3G theorem for polyharmonic Green function on
the unit ball of Rn (n ≥ 2) corresponding to zero Dirichlet boundary conditions.
This enables us to introduce a new class of functions Km,n containing properly
the classical Kato class Kn . We exploit properties of functions belonging to Km,n
to prove an infinite existence result of singular positive solutions for nonlinear
elliptic equation of order 2m.
1. Introduction
In [2], Boggio gave an explicit expression for the Green function Gm,n of (−)m
on the unit ball B of Rn (n ≥ 2) with Dirichlet boundary conditions
u=
∂
∂m−1
u = · · · = m−1 u = 0 on ∂B,
∂ν
∂ν
(1.1)
where ∂/∂ν is the outward normal derivate and m is a positive integer.
In fact, he proved that for each x, y in B, we have
Gm,n (x, y) = km,n |x − y |2m−n
[x,y]/|x− y | 1
m−1
v2 − 1
v n −1
dv,
(1.2)
where km,n is a positive constant and [x, y]2 = |x − y |2 + (1 − |x|2 )(1 − | y |2 ), for
each x, y in B.
Hence, from its expression, it is clear that Gm,n is positive in B2 , which does
not hold for the Green function for the biharmonic or m-polyharmonic operator
for an arbitrary bounded domain (see, e.g., [5]). Only for the case m = 1, we do
not have this restriction.
In [7], using the Boggio formula (1.2), Grunau and Sweers have established
some interesting estimates for the Green function Gm,n in B. In particular, they
Copyright © 2003 Hindawi Publishing Corporation
Abstract and Applied Analysis 2003:12 (2003) 715–741
2000 Mathematics Subject Classification: 34B27, 35J40
URL: http://dx.doi.org/10.1155/S1085337503209039
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Singular solutions for polyharmonic equation
obtained the following inequality called 3G theorem: there exists a constant
am,n > 0 such that for each x, y,z ∈ B,


|x − z|2m−n + |z − y |2m−n ,



 3 3
for 2m < n,
Gm,n (x,z)Gm,n (z, y)
≤ am,n log
+ log
, for 2m = n,

Gm,n (x, y)
|x − z|
|z − y |




1,
for 2m > n.
(1.3)
The Green function for the Laplacian (m = 1) satisfies the above inequality
in an arbitrary bounded C 1,1 domain Ω in Rn . In fact, for the case n ≥ 3, Zhao
proved in [19] the existence of a positive constant Cn such that for each x, y,z in
Ω,
G1,n (x,z)G1,n (z, y)
1
1
≤ Cn
+
.
n
−
2
G1,n (x, y)
|x − z|
| y − z|n−2
(1.4)
Moreover, for the case n = 2, Chung and Zhao showed in [3] the existence of a
positive constant C2 such that for each x, y,z in Ω,
G1,2 (x,z)G1,2 (z, y)
1
≤ C2 max 1,log
G1,2 (x, y)
|x − z|
+ max 1,log
1
| y − z|
.
(1.5)
The 3G theorem related to G1,n has been exploited in the study of functions
belonging to the Kato class Kn (Ω) (see Definition 1.1), which was widely used in
the study of some nonlinear differential equations (see [15, 18]).
More properties pertaining to this class can be found in [1, 3].
Definition 1.1 (see [1, 3]). A Borel measurable function ϕ in Ω belongs to the
Kato class Kn (Ω) if ϕ satisfies the following conditions:
ϕ(y)
d y = 0,
|x − y |n−2
1
log
ϕ(y) d y = 0,
lim sup
α→0 x∈Ω Ω∩B(x,α)
|x − y |
lim sup
α→0
x∈Ω Ω∩B(x,α)
if n ≥ 3,
(1.6)
if n = 2.
The purpose of this paper is two-folded. One is to give a new form of the 3G
theorem to the Green function Gm,n in B 2 which improves (1.3) and enables us
to introduce a new Kato class Km,n := Km,n (B) in the sense of Definition 1.2. The
Imed Bachar et al. 717
second purpose is to investigate the existence of infinitely many singular positive
solutions for the following nonlinear elliptic problem:
∆m u = (−1)m f (·,u) in B \ {0} (in the sense of distributions),
∂
∂m−1
u = · · · = m−1 u = 0 on ∂B,
∂ν
∂ν
u(x) ∼ cρ(x), near x = 0, for any sufficiently small c > 0,
u=
(1.7)
where

1


,


n−2m

|
x
|
 for 2m < n,
,
for 2m = n,
ρ(x) = log





1
|x|
(1.8)
for 2m > n,
1,
and f is required to satisfy suitable assumptions related to the class Km,n which
will be specified later.
The existence of infinitely many singular positive solutions for problem (1.7)
in the case m = 1, for an arbitrary bounded C 1,1 domain Ω in Rn (n ≥ 3), has
been established by Zhang and Zhao in [18] for the special nonlinearity
f (x,t) = p(x)t µ ,
µ > 1,
(1.9)
∈ Kn (Ω).
(1.10)
where the function p satisfies
x −→
p(x)
|x|(n−2)(µ−1)
This result has been recently extended by Mâagli and Zribi in [14], where f
satisfies some appropriate conditions related to the class K1,n (Ω).
Here we extend these results to the high order.
The outline of the paper is as follows. In Section 2, we find again by a simpler argument some estimates on the Green function Gm,n given by Grunau and
Sweers in [7] and we give further ones, including the following:
δ(y)
δ(x)
m

1



n−2m ,


|
x
−
y

 |
Gm,n (x, y) ≤ C log





1,
3
|x − y |
for 2m < n,
, for 2m = n,
for 2m > n.
(1.11)
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Singular solutions for polyharmonic equation
Next, we establish the 3G theorem in this form: there exists Cm,n > 0 such that
for each x, y,z ∈ B,
Gm,n (x,z)Gm,n (z, y)
Gm,n (x, y)
≤ Cm,n
δ(z)
δ(x)
m
δ(z)
Gm,n (x,z) +
δ(y)
m
(1.12)
Gm,n (y,z) ,
which improves (1.3). We note that, for m = 1, (1.12) holds for an arbitrary
bounded domain Ω in Rn . This was proved by Kalton and Verbitsky in [10] for
n ≥ 3 and by Selmi in [16] for the case n = 2.
In Section 3, we define and study some properties of functions belonging to
the class Km,n .
Definition 1.2. A Borel measurable function ϕ in B belongs to the class Km,n if ϕ
satisfies the following condition:
lim sup
α→0
x∈B B ∩B(x,α)
δ(y)
δ(x)
Gm,n (x, y) ϕ(y) d y = 0.
m
(1.13)
In particular, we show that Km,n contains properly K j,n , for 1 ≤ j ≤ m − 1,
which contains properly Kn (B). We close this section by giving a characterization
of the radial functions belonging to the class Km,n .
For the case m = 1, this class has been extensively studied for an arbitrary
bounded C 1,1 domain in Rn , in [14], for n ≥ 3, and in [12, 17] for n = 2. To study
problem (1.7) in Section 4, we assume that f satisfies the following hypotheses:
(H1 ) f is a Borel measurable function on B × (0, ∞), continuous with respect
to the second variable;
(H2 ) | f (x,t)| ≤ tq(x,t), where q is a nonnegative Borel measurable function
in B × (0, ∞), such that the function t → q(x,t) is nondecreasing on
(0, ∞) and limt→0 q(x,t) = 0;
(H3 ) the function g, defined on B by g(x) = q(x,Gm,n (x,0)), belongs to the
class Km,n .
We point out that in the case m = 1 and f (x,t) = p(x)t µ , the assumption
(1.10) implies (H3 ).
In order to simplify our statements, we define some convenient notation.
Notation. (i) We denote B = {x ∈ Rn ; |x| < 1} with n ≥ 2.
(ii) We denote s ∧ t = min(s,t) and s ∨ t = max(s,t) for s,t ∈ R.
(iii) For x, y ∈ B,
[x, y]2 = |x − y |2 + 1 − |x|2 1 − | y |2 ,
δ(x) = 1 − |x|,
2
2
2
θ(x, y) = [x, y] − |x − y | = 1 − |x|
2
1 − | y| .
(1.14)
Imed Bachar et al. 719
Note that [x, y]2 ≥ 1 + |x|2 | y |2 − 2|x|| y | = (1 − |x|| y |)2 . So we have
δ(x) ≤ [x, y],
δ(y) ≤ [x, y].
(1.15)
(iv) Let f and g be positive functions on a set S.
We call f ∼ g if there is c > 0 such that
1
g(x) ≤ f (x) ≤ cg(x) ∀x ∈ S.
c
(1.16)
We call f g if there is c > 0 such that
f (x) ≤ cg(x) ∀x ∈ S.
(1.17)
The following properties will be used several times:
(i) for s,t ≥ 0, we have
st
,
s+t
(s + t) p ∼ s p + t p ,
s∧t ∼
(1.18)
p ∈ R+ ;
(1.19)
(ii) let λ, µ > 0 and 0 < γ ≤ 1, then we have
1 − tλ ∼ 1 − tµ
log(1 + t) t
γ
for t ∈ [0,1],
(1.20)
for t ≥ 0,
(1.21)
log(1 + λt) ∼ log(1 + µt) for t ≥ 0,
log 1 + t
λ
λ
∼ t log(2 + t)
(1.22)
for t ∈ [0,1];
(1.23)
(iii) on B2 (i.e., (x, y) ∈ B2 ), we have
θ(x, y) ∼ δ(x)δ(y),
(1.24)
[x, y]2 ∼ |x − y |2 + δ(x)δ(y).
(1.25)
2. Inequalities for the Green function
We first find another expression of Gm,n given by Hayman and Korenblum in
[8], which will be used later.
Proposition 2.1. The Green function Gm,n satisfies
Gm,n (x, y) = αm,n
m+k
∞
Γ(n/2 + k) θ(x, y)
k =0
where αm,n is some fixed positive constant.
(k + m)![x, y]n+2k
,
(2.1)
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Singular solutions for polyharmonic equation
Proof. Using the transformation v2 = 1 + (θ(x, y)/ |x − y |2 )(1 − t) in (1.2), Gm,n
becomes
m 1
θ(x, y)
k
Gm,n (x, y) = m,n
2
[x, y]n
0
(1 − t)m−1
n/2 dt.
1 − t θ(x, y)/[x, y]2
(2.2)
Since 0 < θ(x, y)/[x, y]2 ≤ 1, and for each t ∈ [0,1[, we have
(1 − t)−n/2 =
∞
Γ(n/2 + k)
k =0
k!Γ(n/2)
tk ;
(2.3)
it follows that
m+k
∞
k Γ(n/2 + k) θ(x, y)
B(k + 1,m),
Gm,n (x, y) = m,n
2 k=0 k!Γ(n/2) [x, y]n+2k
where B(k + 1,m) :=
That is,
1
0
(2.4)
t k (1 − t)m−1 dt = k!(m − 1)!/(k + m)!.
Gm,n (x, y) = αm,n
m+k
∞
Γ((n/2) + k) θ(x, y)
(2.5)
[x, y]n+2k
(k + m)!
k =0
with αm,n > 0.
Moreover, from formula (1.2), we may prove, by simpler argument, the following estimates on Gm,n given in [7].
Proposition 2.2. On B2 , the following estimates hold:
(i) for 2m < n,
2m−n
Gm,n (x, y) ∼ |x − y |
m δ(x)δ(y)
1∧
|x − y |2m
;
(2.6)
(ii) for 2m = n,
m δ(x)δ(y)
Gm,n (x, y) ∼ log 1 +
|x − y |2m
;
(2.7)
(iii) for 2m > n,
m−n/2
Gm,n (x, y) ∼ δ(x)δ(y)
n/2 δ(x)δ(y)
1∧
|x − y |n
.
(2.8)
Proof. Using in (1.2) the transformation t = (v2 − 1)m , we obtain the following
expression for Gm,n :
Gm,n (x, y) = C |x − y |2m−n
(θ(x,y)/|x− y |2 )m
0
dt
n/2 .
t 1/m + 1
(2.9)
Imed Bachar et al. 721
Now, from (1.19) we have
Gm,n (x, y) ∼ |x − y |2m−n
(θ(x,y))m /|x− y |2m
dt
.
t n/2m + 1
0
(2.10)
Next, we distinguish the following cases.
Case 1 (2m = n). It follows from (2.10), (1.22), and (1.24) that
m θ(x, y)
Gm,n (x, y) ∼ log 1 +
|x − y |2m
∼ log 1 +
m δ(x)δ(y)
|x − y |2m
(2.11)
.
Case 2 (2m < n). Using the fact that for each a > 0 and λ > 1, we have
a
0
1
dt ∼ 1 ∧ a,
tλ + 1
(2.12)
hence, we deduce from (2.10) and (1.24) that
2m−n
Gm,n (x, y) ∼ |x − y |
2m−n
∼ |x − y |
m θ(x, y)
1∧
|x − y |2m
m δ(x)δ(y)
1∧
|x − y |2m
(2.13)
.
Case 3 (2m > n). We recall that 0 < θ(x, y)/[x, y]2 ≤ 1, which yields
1
0
(1 − t)m−1
n/2 dt ∼ 1.
1 − t θ(x, y)/[x, y]2
(2.14)
This implies, with (2.2), that
m
θ(x, y)
,
Gm,n (x, y) ∼
[x, y]n
(2.15)
which, together with (1.24), (1.18), and (1.19), gives that
m−n/2
Gm,n (x, y) ∼ δ(x)δ(y)
1∧
n/2 δ(x)δ(y)
|x − y |n
.
(2.16)
Corollary 2.3. On B2 , the following estimates hold:
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Singular solutions for polyharmonic equation
(i) if 2m < n,
m
Gm,n (x, y) ∼
∼
∼
δ(x)δ(y)
m
|x − y |n−2m |x − y |2 + δ(x)δ(y)
m
δ(x)δ(y)
(2.17)
|x − y |n−2m [x, y]2m
1
|x − y |n−2m
−
2m
|x − y |
1
m (n−2m)/2m ;
+ δ(x)δ(y)
(ii) if 2m = n,
m m
δ(x)δ(y)
Gm,n (x, y) ∼ 1 ∧
|x − y |2m
δ(x)δ(y)
log 2 +
|x − y |2
δ(x)δ(y)
δ(x)δ(y)
m log 2 +
∼
2
|x − y |2
|x − y | + δ(x)δ(y)
m
δ(x)δ(y)
∼
[x, y]2m
(2.18)
[x, y]2
log 1 +
;
|x − y |2
(iii) if 2m > n,
m
δ(x)δ(y)
Gm,n (x, y) ∼ n/2
|x − y |2 + δ(x)δ(y)
m
δ(x)δ(y)
∼
[x, y]n
(2.19)
.
Proof. The proof follows immediately from Proposition 2.2 and the statements
(1.18), (1.19), (1.20), (1.22), (1.23), (1.24), and (1.25).
From the above estimates, we derive some inequalities for the Green function
Gm,n including (1.11), which will be done in the following corollaries.
Corollary 2.4. On B2 , the following estimates hold:
δ(y)
δ(x)
m

1


,



y |n−2m

 |x −
Gm,n (x, y) log






3
|x − y |
for 2m < n,
, for 2m = n,
(2.20)
for 2m > n.
1,
Proof. Using Corollary 2.3 and inequalities (1.15), we deduce that
(i) if 2m < n,
δ(y)
δ(x)
m
Gm,n (x, y) 2m
δ(y)
1
1
≤
;
|x − y |n−2m [x, y]2m
|x − y |n−2m
(2.21)
Imed Bachar et al. 723
(ii) if 2m = n,
δ(y)
δ(x)
m
Gm,n (x, y) log 1 +
[x, y]2
|x − y |2
2m
δ(y)
3
≤ log
;
[x, y]2m
|x − y |
(2.22)
(iii) if 2m > n,
δ(y)
δ(x)
m
Gm,n (x, y) 2m
δ(y)
≤ 1.
[x, y]n
(2.23)
Corollary 2.5. For each x, y ∈ B such that |x − y | ≥ r,
m
δ(x)δ(y)
rn
Gm,n (x, y) .
(2.24)
Moreover, on B 2 , the following estimates hold:
m
Gm,n (x, y),
m m
Gm,n (x, y) δ(x) ∧ δ(y)
m m
δ(x) ∧ δ(y)
Gm,n (x, y) |x − y |n−m
δ(x)δ(y)
(2.25)
if m ≥ n,
(2.26)
if 1 ≤ m < n.
(2.27)
Proof. Assertions (2.24) and (2.25) are obviously obtained using the estimates
in Corollary 2.3 and the fact that |x − y | ≤ [x, y] 1.
Now, if m ≥ n, then we deduce from Corollary 2.3 and (1.15) that
m
δ(x)δ(y)
Gm,n (x, y) ∼
[x, y]n
m m
δ(x) ∧ δ(y) .
(2.28)
Then (2.26) holds.
To prove (2.27), we suppose that 1 ≤ m < n. So we obtain, from Corollary 2.3,
inequalities (1.15), and |x − y | ≤ [x, y] that
(i) if 2m < n, then we have
m
δ(x)δ(y)
Gm,n (x, y) ∼
|x − y |n−2m [x, y]2m
≤
m m
δ(y)
δ(x)
|x − y |n−m [x, y]m
m
δ(x)
≤
;
|x − y |n−m
(2.29)
724
Singular solutions for polyharmonic equation
(ii) if 2m = n, then using further inequality (1.21), we deduce that
Gm,n (x, y) ∼ log 1 +
[x, y]2
|x − y |2
m
δ(x)δ(y)
[x, y]2m
m
[x, y] δ(x)δ(y)
|x − y | [x, y]2m
m m
(2.30)
m
δ(y)
δ(x)
≤
|x − y |m [x, y]m
δ(x)
;
|x − y |m
≤
(iii) if 2m > n, then we have
m
δ(x)δ(y)
[x, y]n
Gm,n (x, y) ∼
≤
m m
m
δ(y)
δ(x)
δ(x)
≤
.
|x − y |n−m [x, y]m
|x − y |n−m
(2.31)
Hence interchanging the roles of x and y, (2.27) is proved.
In the sequel, for a nonnegative measurable function f on B, we put
Vm,n f (x) =
B
for x ∈ B.
Gm,n (x, y) f (y)d y
(2.32)
Remark 2.6. Let m ≥ n. Then there exists a positive constant C1 such that, for
each f ∈ L1+ (B) and x ∈ B, we have
1
C1
δ(y)
B
m
f (y)d y
δ(x)
m
m
≤ Vm,n f (x) ≤ C1 f 1 δ(x) .
(2.33)
In particular, we have Vm,n 1(x) ∼ (δ(x))m .
Moreover, let 1 ≤ m < n. Then there exists a positive constant C2 such that for
p
each f ∈ L+ (B) with p > n/m and x ∈ B, we have
1
C2
B
δ(y)
m
f (y)d y
δ(x)
m
m
≤ Vm,n f (x) ≤ C2 f p δ(x) .
(2.34)
Indeed, (2.33) holds by (2.25) and (2.26). To prove (2.34), we use (2.25) and
(2.27) and we apply the Hölder inequality, so we obtain that, for x ∈ B,
B
δ(y)
m
f (y)d y
δ(x)
m
Vm,n f (x)
m
δ(x) f p
B
dy
|x − y |(n−m)p/(p−1)
(p−1)/ p
.
(2.35)
Imed Bachar et al. 725
Now, for each x ∈ B, we have
dy
B
|x − y |(n−m)p/(p−1)
≤
dξ
B(0,2)
|ξ |(n−m)p/(p−1)
,
(2.36)
and this last integral is finite if and only if p > n/m, which gives (2.34).
Next, we aim to prove inequality (1.12). So, we need the following key lemma.
Lemma 2.7 (see [11, 13]). Let x, y ∈ B. Then the following properties are satisfied:
√
(1) if δ(x)δ(y) ≤ |x − y |2 , then (δ(x) √
∨ δ(y)) ≤ (( 5 + 1)/2)|x − y |;
√
(2) if |x − y |2 ≤ δ(x)δ(y), then ((3 − 5)/2)δ(x) ≤ δ(y) ≤ ((3 + 5)/2)δ(x).
Proof. (1) We may assume that (δ(x) ∨ δ(y)) = δ(y). Then the inequalities δ(y) ≤
δ(x) + |x − y | and δ(x)δ(y) ≤ |x − y |2 imply that
2
δ(y) − δ(y)|x − y | − |x − y |2 ≤ 0,
(2.37)
that is,
√
δ(y) +
√
5−1
|x − y |
2
5+1
|x − y | ≤ 0.
2
δ(y) −
(2.38)
It follows that
δ(x) ∨ δ(y) ≤
√
5+1
|x − y |.
2
(2.39)
(2) For each z ∈ ∂B, we have | y − z| ≤ |x − y | + |x − z| and since |x − y |2 ≤
δ(x)δ(y), we obtain
| y − z| ≤ δ(x)δ(y) + |x − z| ≤ |x − z|| y − z| + |x − z|,
(2.40)
√
√
5−1 5+1 | y − z| +
|x − z|
| y − z| −
|x − z| ≤ 0.
(2.41)
that is,
2
2
It follows that
√ 3+ 5
| y − z| ≤
|x − z|.
2
(2.42)
Thus, interchanging the roles of x and y, we have
√ √ 3− 5
3+ 5
|x − z| ≤ | y − z| ≤
|x − z|,
2
2
(2.43)
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Singular solutions for polyharmonic equation
which gives
√ √ 3− 5
3+ 5
δ(x) ≤ δ(y) ≤
δ(x).
2
2
(2.44)
Theorem 2.8 (3G theorem). There exists a constant Cm,n > 0 such that, for each
x, y,z ∈ B,
Gm,n (x,z)Gm,n (z, y)
Gm,n (x, y)
≤ Cm,n
δ(z)
δ(x)
m
Gm,n (x,z) +
δ(z)
δ(y)
m
(2.45)
Gm,n (y,z) .
Proof. To prove the inequality, we denote A(x, y) := (δ(x)δ(y))m /Gm,n (x, y) and
we claim that A is a quasimetric, that is, for each x, y,z ∈ B,
A(x, y) A(x,z) + A(y,z).
(2.46)
To show the claim, we separate the proof into three cases.
Case 1. For 2m < n, using Proposition 2.2, we have
m
A(x, y) ∼ |x − y |n−2m |x − y |2 ∨ δ(x)δ(y)
.
(2.47)
We distinguish the following subcases:
(i) if δ(x)δ(y) ≤ |x − y |2 , then we have
A(x, y) ∼ |x − y |n |x − z|n + | y − z|n A(x,z) + A(y,z);
(2.48)
(ii) the inequality |x − y |2 ≤ δ(x)δ(y) implies, from Lemma 2.7, that δ(x) ∼
δ(y). So we deduce the following:
(a) if |x − z|2 ≤ δ(x)δ(z) or | y − z|2 ≤ δ(y)δ(z), then it follows from Lemma
2.7 that δ(x) ∼ δ(y) ∼ δ(z). Hence,
m
A(x, y) ∼ |x − y |n−2m δ(x)δ(y)
m δ(x)δ(y)
|x − z|n−2m + | y − z|n−2m
m
m
|x − z|n−2m δ(x)δ(z) + | y − z|n−2m δ(y)δ(z)
(2.49)
A(x,z) + A(y,z);
(b) if |x − z|2 ≥ δ(x)δ(z) and | y − z|2 ≥ δ(y)δ(z), then using Lemma 2.7,
we have
δ(x) ∨ δ(z) |x − z|,
δ(y) ∨ δ(z) | y − z|.
(2.50)
Imed Bachar et al. 727
So, we have
m
A(x, y) ∼ |x − y |n−2m δ(x)δ(y)
m
|x − z|n−2m + | y − z|n−2m δ(x)δ(y)
2m
2m
|x − z|n−2m δ(x)
+ | y − z|n−2m δ(y)
(2.51)
|x − z|n + | y − z|n
A(x,z) + A(y,z).
Case 2. For 2m = n, using Proposition 2.2, we have
A(x, y) ∼
m
δ(x)δ(y)
m
.
log 1 + δ(x)δ(y) / |x − y |2m
(2.52)
Then, since for each t ≥ 0,
t
≤ log(1 + t) ≤ t,
1+t
(2.53)
m
|x − y |2m A(x, y) |x − y |2m + δ(x)δ(y) .
(2.54)
we deduce that
So we distinguish the following subcases:
(i) if δ(x)δ(y) ≤ |x − y |2 , then by (1.19), we have
A(x, y) |x − y |2m |x − z|2m + | y − z|2m A(x,z) + A(y,z);
(2.55)
(ii) if |x − y |2 ≤ δ(x)δ(y), it follows by Lemma 2.7 that δ(x) ∼ δ(y).
So, we distinguish the following two subcases:
(a) if |x − z|2 ≤ δ(x)δ(z) or | y − z|2 ≤ δ(y)δ(z), so from Lemma 2.7, we
deduce that δ(x) ∼ δ(y) ∼ δ(z).
Now, since
|x − y |2m |x − z|2m + | y − z|2m |x − z|2m ∨ | y − z|2m ,
(2.56)
then we obtain that
m δ(x)δ(z)
log 1 +
|x − z|2m
m δ(y)δ(z)
∧ log 1 +
| y − z|2m
m δ(x)δ(y)
log 1 +
|x − y |2m
(2.57)
,
which, together with (2.52), implies that
A(x, y) A(x,z) + A(y,z);
(2.58)
728
Singular solutions for polyharmonic equation
(b) if |x − z|2 ≥ δ(x)δ(z) and | y − z|2 ≥ δ(y)δ(z), then by Lemma 2.7, it
follows that
δ(x) ∨ δ(z) |x − z|,
δ(y) ∨ δ(z) | y − z|.
(2.59)
Hence, by (2.54), we have
m
A(x, y) δ(x)δ(y)
2m 2m
δ(x)
+ δ(y)
(2.60)
|x − z|2m + | y − z|2m
A(x,z) + A(y,z).
Case 3. For 2m > n, from Proposition 2.2, we have
n/2
A(x, y) ∼ |x − y |2 ∨ δ(x)δ(y)
.
(2.61)
Then the result holds by arguments similar to that of Case 2(i).
3. The Kato class Km,n
In this section, we will study properties of functions belonging to the class Km,n .
We first compare the classes K j,n for j ≥ 1.
Proposition 3.1. For each m ≥ 1, the following estimate is satisfied on B2 :
δ(y)
δ(x)
m
Gm,n (x, y) δ(y)
2(m−1) δ(y)
G1,n (x, y).
δ(x)
(3.1)
In particular, K1,n ⊂ (δ(·))2(m−1) Km,n .
Proof. Using (1.2), we have
Gm,n (x, y) |x − y |2m−n
[x, y]2
−1
|x − y |2
m−1 [x,y]/ |x− y |
1
dv
.
v n −1
(3.2)
Now, we remark by (1.25) that
[x, y]2
δ(x)δ(y)
−1 ∼
.
|x − y |2
|x − y |2
(3.3)
So we deduce that
m−1
Gm,n (x, y) δ(x)δ(y)
G1,n (x, y),
which implies (3.1). The proof is complete by (1.13).
(3.4)
Imed Bachar et al. 729
Remark 3.2. Let j,m ∈ N such that 1 ≤ j < m, then we have
Kn (B) ⊂ K j,n ⊂ Km,n .
(3.5)
Indeed, by a similar argument as above, we prove that, on B2 ,
δ(y)
δ(x)
m
Gm,n (x, y) δ(y)
2(m− j) δ(y) j
δ(x)
G j,n (x, y),
(3.6)
which implies that K j,n ⊂ Km,n . The first inclusion in (3.5) holds by putting m =
1 in Corollary 2.4.
Lemma 3.3. Let ϕ be a function in Km,n . Then the function
x −→ δ(x)
2m
ϕ(x)
(3.7)
is in L1 (B).
Proof. Let ϕ ∈ Km,n , then by (1.13), there exists α > 0 such that for each x ∈ B,
B(x,α)∩B
δ(y)
δ(x)
m
Gm,n (x, y)ϕ(y)d y ≤ 1.
(3.8)
Let x1 ,...,x p be in B such that B ⊂ ∪1≤i≤ p B(xi ,α). Then by (2.25), there exists
C > 0 such that for all i ∈ {1,..., p} and y ∈ B(xi ,α) ∩ B, we have
δ(y)
2m
≤C
δ(y)
δ(xi )
m
Gm,n (xi , y).
(3.9)
Hence, we have
B
δ(y)
2m ϕ(y)d y ≤ C
1≤i≤ p B xi ,α ∩B
δ(y)
δ xi
m
Gm,n xi , y ϕ(y)d y
≤ C p < ∞.
(3.10)
This completes the proof.
In the sequel, we use the notation
ϕB := sup
x ∈B B
δ(y)
δ(x)
m
Gm,n (x, y)ϕ(y)d y.
Proposition 3.4. Let ϕ be a function in Km,n , then ϕB < ∞.
(3.11)
730
Singular solutions for polyharmonic equation
Proof. Let ϕ ∈ Km,n and α > 0. Then we have
B
δ(y)
δ(x)
m
Gm,n (x, y)ϕ(y)d y
δ(y) m
≤
Gm,n (x, y)ϕ(y)d y
δ(x)
B ∩|x− y |≤α
δ(y) m
Gm,n (x, y)ϕ(y)d y.
+
B ∩|x− y |≥α δ(x)
(3.12)
Now, since by (2.24), we have
B ∩|x− y |≥α
δ(y)
δ(x)
m
1
Gm,n (x, y)ϕ(y)d y αn
B
δ(y)
2m ϕ(y)d y,
(3.13)
then the result follows from (1.13) and Lemma 3.3.
Proposition 3.5. There exists a constant C > 0 such that, for all ϕ ∈ Km,n and h
a nonnegative harmonic function in B,
B
Gm,n (x, y) δ(y)
m−1
h(y)ϕ(y)d y ≤ C ϕB δ(x)
m−1
h(x)
(3.14)
for all x in B.
Proof. Let h be a nonnegative harmonic function in B. So by Herglotz representation theorem (see [9, page 29]), there exists a nonnegative measure µ on ∂B
such that
h(y) =
∂B
P(y,ξ)µ(dξ),
(3.15)
where P(y,ξ) = (1 − | y |2 )/ | y − ξ |n , for y ∈ B and ξ ∈ ∂B. So we need only to
verify (3.14) for h(y) = P(y,ξ) uniformly in ξ ∈ ∂B.
By (2.1) we have for each x, y ∈ B,
Gm,n (x, y) = αm,n
m
θ(x, y) 1 + o 1 − | y |2 .
n
[x, y]
(3.16)
Hence, for x, y,z in B,
m
1 − | y |2 [x,z]n Gm,n (y,z)
m
=
1
+
o
1
− |z|2 ,
Gm,n (x,z)
1 − |x|2 [y,z]n
(3.17)
Imed Bachar et al. 731
which implies that
m
Gm,n (y,z)
1 − | y |2 |x − ξ |n
δ(y)
m
=
∼
lim
δ(x)
z→ξ Gm,n (x,z)
1 − |x|2 | y − ξ |n
m−1
P(y,ξ)
.
P(x,ξ)
(3.18)
Thus by Fatou’s lemma and (1.12), we deduce that
B
m−1
P(y,ξ) ϕ(y)d y
P(x,ξ)
Gm,n (y,z) ϕ(y)d y
liminf Gm,n (x, y)
Gm,n (x,z)
z→ξ
B
δ(y) m
liminf
Gm,n (x, y)ϕ(y)d y
z→ξ
B δ(x)
δ(y) m
Gm,n (z, y)ϕ(y)d y
+
B δ(z)
ϕB ,
Gm,n (x, y)
δ(y)
δ(x)
(3.19)
which completes the proof.
Corollary 3.6. Let ϕ be in Km,n . Then
sup
x ∈B
B
Gm,n (x, y) δ(y)
m−1 ϕ(y)d y < ∞.
(3.20)
Moreover, the function x → (δ(x))2m−1 ϕ(x) is in L1 (B).
Proof. Put h ≡ 1 in (3.14) and using Proposition 3.4, we get (3.20).
Moreover, by (2.25), it follows that
B
δ(y)
2m−1 ϕ(y)d y B
Gm,n (0, y) δ(y)
m−1 ϕ(y)d y.
(3.21)
Hence the result follows from (3.20).
Remark 3.7. We recall (see [1]) that for m = 1 and n ≥ 3, a radial function ϕ is
in the classical Kato class Kn (B) if and only if
1
0
r ϕ(r)dr < ∞.
(3.22)
Similarly, we will give in the sequel a characterization of the radial functions
belonging to Km,n , which asserts, in particular, that inclusions (3.5) are proper.
More precisely, we will prove in the next proposition that a radial function ϕ is
in Km,n if and only if (3.20) is satisfied.
732
Singular solutions for polyharmonic equation
Proposition 3.8. Let ϕ be a radial function in B, then the following assertions are
equivalent:
(1) ϕ ∈ Km,n
;
(2) supx∈B B Gm,n (x, y)(δ(y))m−1 |ϕ(y)|d y < ∞;
(3) for 2m < n,
1
0
r 2m−1 (1 − r)2m−1 ϕ(r)dr < ∞.
(3.23)
For 2m = n,
1
r n−1 (1 − r)n−2 log
0
1 ϕ(r)dr < ∞.
(3.24)
r
For 2m > n,
1
0
r n−1 (1 − r)2m−1 ϕ(r)dr < ∞.
Proof. Since the function x →
note that t = |x| and
Sn−1 Gm,n (x,rω)dσ(ω)
(3.25)
is radial in B, then we de-
ψm,n (t,r) =
Sn−1
Gm,n (x,rω)dσ(ω),
(3.26)
where σ is the normalized measure on the unit sphere Sn−1 of Rn .
Now, using Corollary 2.3 and the fact that for each y ∈ B, [0, y] = 1, we deduce that

2m−n (1 − r)m ,


r


ψm,n (0,r) ∼ (1 − r)m log 1 + 1 ∼ (1 − r)m−1 log 1 ,

r2
r



m
(1 − r) ,
for 2m < n,
for 2m = n, (3.27)
for 2m > n.
So, assertion (3) is equivalent to
1
(3 ) 0 r n−1 (1 − r)m−1 ψm,n (0,r)|ϕ(r)|dr < ∞.
We now prove the equivalences.
(1)⇒(2) follows from Corollary 3.6.
(2)⇔(3 ). By virtue of [4, Theorem 2.4], we have that t → ψm,n (t,r) is a nonincreasing map on [0,1], so that
sup
x ∈B B
Gm,n (x, y) δ(y)
= sup
1
t ∈[0,1] 0
=
1
0
m−1 ϕ(y)d y
r n−1 (1 − r)m−1 ψm,n (t,r)ϕ(r)dr
r n−1 (1 − r)m−1 ψm,n (0,r)ϕ(r)dr.
(3.28)
Imed Bachar et al. 733
(3 )⇒(1). Let 0 < α < 1/4, then we have
sup
x∈B B ∩B(x,α)
≤ sup
δ(y)
δ(x)
m
(t+α)∧1
r n −1
0≤t ≤1 (t −α)∨0
(t+α)∧1
≤ sup
Gm,n (x, y)ϕ(y)d y
(1 − r)m
ψm,n (t,r)ϕ(r)dr
(1 − t)m
r n −1
0≤t ≤1/2 (t −α)∨0
(t+α)∧1
+ sup
1/2≤t ≤1 (t −α)∨0
(1 − r)m
ψm,n (t,r)ϕ(r)dr
(1 − t)m
r n −1
(3.29)
(1 − r)m
ψm,n (t,r)ϕ(r)dr
m
(1 − t)
= I1 + I2 .
Using [4, Theorem 2.4], we have
I1 ≤ sup
(t+α)∧1
0≤t ≤1/2 (t −α)∨0
(t+α)∧1
sup
0≤t ≤1/2 (t −α)∨0
r n −1
r
n −1
(1 − r)m
ψm,n (0,r)ϕ(r)dr
(1 − t)m
(1 − r)
m−1
ψm,n (0,r)ϕ(r)dr.
(3.30)
On the other hand, by (3.1), we have
I2 ≤ sup
(t+α)∧1
1/2≤t ≤1 (t −α)∨0
r n −1
(1 − r)2m−1
ψ1,n (t,r)ϕ(r)dr.
(1 − t)
(3.31)
Now, by elementary calculus, we obtain that

1



(t ∨ r)2−n 1 − (t ∨ r)n−2 ,

n−2
ψ1,n (t,r) =  1


,
log
t∨r
for n ≥ 3,
(3.32)
for n = 2.
So, using (1.20) and the fact that log(1/s) (1 − s) for s ≥ 1/2, we have for each
n ≥ 2 and t ≥ 1/2,
ψ1,n (t,r) (1 − t ∨ r).
(3.33)
Hence, from (3.27), we have
I2 sup
(t+α)∧1
1/2≤t ≤1 (t −α)∨0
(t+α)∧1
sup
1/2≤t ≤1 (t −α)∨0
Thus, I1 + I2 sup0≤t≤1
(t+α)∧1
(t −α)∨0
r n−1 (1 − r)2m−1
r
n −1
(1 − r)
m−1
(1 − t ∨ r) ϕ(r)dr
1−t
ψm,n (0,r)ϕ(r)dr.
r n−1 (1 − r)m−1 ψm,n (0,r)|ϕ(r)|dr.
(3.34)
734
Singular solutions for polyharmonic equation
s
Let φ(s) = 0 r n−1 (1 − r)m−1 ψm,n (0,r)|ϕ(r)|dr for s ∈ [0,1].
Then using (3 ), we deduce that φ is a continuous function on [0,1], which
implies that
(t+α)∧1
r n−1 (1 − r)m−1 ψm,n (0,r)ϕ(r)dr
(t −α)∨0
= φ (t + α) ∧ 1 − φ (t − α) ∨ 0
(3.35)
converges to zero as α → 0 uniformly for t ∈ [0,1]. So, limα→0 (I1 + I2 ) = 0, that
is, ϕ ∈ Km,n .
Example 3.9. let q be the function defined in B by
1
q(x) = λ .
δ(x)
(3.36)
By Proposition 3.8, q ∈ Km,n if and only if λ < 2m and Vm,n q is bounded if and
only if λ < m + 1. In fact, we give in the next proposition more precise estimates
on the m-potential Vm,n q.
Proposition 3.10. On B, the following estimates hold:
(i) (δ(x))m Vm,n q(x) (δ(x))2m−λ if m < λ < m + 1;
(ii) (δ(x))m Vm,n q(x) (δ(x))m log(2/δ(x)) if λ = m;
(iii) Vm,n q(x) ∼ (δ(x))m if λ < m.
Proof. Let λ < m + 1. Then from (2.25), we have
δ(x)
m
B
dy
δ(y)
λ−m Vm,n q(x),
(3.37)
which implies the lower estimates.
For the upper estimates, we have, from (3.1),
Vm,n q(x) B
δ(x)
δ(x)
m−1 m−1
δ(y)
1
0
m−1
G1,n (x, y)q(y)d y
r n −1
ψ1,n |x|,r dr.
(1 − r)λ+1−m
(3.38)
On the other hand, using (1.20) and the inequality t log(1/t) ≤ (1 − t), for t ∈
[0,1], we deduce from (3.32) that r n−1 ψ1,n (|x|,r) (1 − |x| ∨ r) for each n ≥ 2.
Imed Bachar et al. 735
This implies that
m−1
1
1 − |x| ∨ r
dr
λ+1−m
0 (1 − r)
m |x|
m−1 1
dr
dr
δ(x)
+
δ(x)
λ+1−m
λ −m
0 (1 − r)
|x| (1 − r)
= I1 + I2 .
Vm,n q(x) δ(x)
(3.39)
So, by elementary calculus, we obtain that

m−λ


δ(x)
,



m
2
,
I1 δ(x) log
δ(x)



1,
I2 δ(x)
2m−λ
if m < λ < m + 1,
if λ = m,
if λ < m,
(3.40)
.
This completes the proof.
Remark 3.11. By Proposition 3.10, we find again the result of Gilbarg and
Trudinger in [6, Theorem 4.9] for the case m = 1 and 1 < λ < 2.
4. Positive singular solutions of the equation ∆m u = (−1)m f (·,u)
In this section, we are interested in the existence of positive singular solutions for
problem (1.7). We present in the next theorem the main result of this section.
Theorem 4.1. Assume (H1 ), (H2 ), and (H3 ). Then problem (1.7) has infinitely
many solutions. More precisely, there exists b0 > 0 such that for each b ∈ (0,b0 ],
there exists a solution u of (1.7) continuous on B \{0} and satisfying for all x ∈ B,
3b
b
Gm,n (x,0) ≤ u(x) ≤ Gm,n (x,0)
2
2
(4.1)
and, for 2m ≤ n,
lim
|x|→0
u(x)
= b.
Gm,n (x,0)
(4.2)
For the proof, we need the following lemmas.
Lemma 4.2. Let ϕ ∈ Km,n and x0 ∈ B. Then
1
lim sup
α→0 x,z∈B Gm,n (x,z)
B ∩B(x0 ,α)
Gm,n (x, y)Gm,n (y,z)ϕ(y)d y = 0.
(4.3)
736
Singular solutions for polyharmonic equation
Proof. Let ε > 0. Then by (1.13), there exists r > 0 such that
sup
δ(y)
δ(ξ)
B ∩B(ξ,r)
ξ ∈B
m
Gm,n (ξ, y)ϕ(y)d y ≤ ε.
(4.4)
Let α > 0. Then it follows, from Theorem 2.8, that for each x,z ∈ B,
1
Gm,n (x,z)
B ∩B(x0 ,α)
≤ Cm,n
B ∩B(x0 ,α)
≤ 2Cm,n sup
δ(y)
δ(x)
B ∩B(x0 ,α)
ξ ∈B
Gm,n (x, y)Gm,n (y,z)ϕ(y)d y
m
δ(y)
Gm,n (x, y) +
δ(z)
δ(y)
δ(ξ)
m
m
Gm,n (y,z) ϕ(y)d y
Gm,n (ξ, y)ϕ(y)d y.
(4.5)
On the other hand, by (2.24), we have
δ(y)
δ(x)
B ∩B(x0 ,α)
≤
m
B ∩(|x− y |≤r)
+
ξ ∈B
+
δ(y)
δ(x)
m
B ∩B(ξ,r)
δ(y)
δ(ξ)
B ∩B(x0 ,α)
δ(y)
Gm,n (x, y)ϕ(y)d y
B ∩B(x0 ,α)∩(|x− y |≥r)
sup
Gm,n (x, y)ϕ(y)d y
m
δ(y)
δ(x)
m
Gm,n (x, y)ϕ(y)d y
(4.6)
Gm,n (ξ, y)ϕ(y)d y
2m ϕ(y)d y.
Now, using Lemma 3.3 and (4.4), the result holds by letting α → 0.
Put F := {ω ∈ C + (B) : ω∞ ≤ 1}, where · ∞ is the uniform norm. So we
have the following result.
Lemma 4.3. Assume (H1 ), (H2 ), and (H3 ). Define the operator T on F by
Tω(x) =
1
Gm,n (x,0)
B
Gm,n (x, y) f y,ω(y)Gm,n (y,0) d y,
x ∈ B.
(4.7)
Then the family of functions T(F) is relatively compact in C(B).
Proof. By (H2 ), we have for all ω ∈ F,
Tω(x) ≤
1
Gm,n (x,0)
B
Gm,n (x, y)Gm,n (y,0)g(y)d y.
(4.8)
Imed Bachar et al. 737
Since g(x) = q(x,Gm,n (x,0)) ∈ Km,n , then, by Theorem 2.8, we deduce that
Tω∞ ≤ 2Cm,n sup
B
ξ ∈B
δ(y)
δ(ξ)
m
Gm,n (ξ, y)g(y)d y
(4.9)
g B .
Hence, the family T(F) is uniformly bounded. Now, we will prove the equicontinuity of T(F) in B. Let x0 ∈ B and α > 0. Let x,x ∈ B(x0 ,α) ∩ B and ω ∈ F,
then
Tω(x) − Tω(x )
Gm,n (x, y) Gm,n (x , y) Gm,n (y,0)g(y)d y
−
≤ Gm,n (x ,0) B Gm,n (x,0)
1
≤ 2sup
Gm,n (ξ, y)Gm,n (y,0)g(y)d y
ξ ∈B
Gm,n (ξ,0)
+ 2sup
+
ξ ∈B
B ∩B(0,2α)
1
Gm,n (ξ,0)
B ∩B(x0 ,2α)
B ∩B c (0,2α)∩B c (x0 ,2α)
(4.10)
Gm,n (ξ, y)Gm,n (y,0)g(y)d y
Gm,n (x, y) Gm,n (x , y) −
G (x,0) G (x ,0) Gm,n (y,0)g(y)d y.
m,n
m,n
If |x0 − y | ≥ 2α, then |x − y | ≥ α and |x − y | ≥ α. So (1.12) and (2.24) imply
that, for all x ∈ B(x0 ,α) ∩ B and y ∈ Ω := Bc (0,2α) ∩ Bc (x0 ,2α) ∩ B,
2m
Gm,n (x, y)
Gm,n (y,0) δ(y) .
Gm,n (x,0)
(4.11)
Moreover, using (3.18), we deduce, when y ∈ Ω, that the function x → Gm,n (x, y)/
Gm,n (x,0) is continuous in B(x0 ,α) ∩ B. Then, by Lemma 3.3 and the dominated
convergence theorem, we obtain that
Gm,n (x, y) Gm,n (x , y) −
G (x,0) G (x ,0) Gm,n (y,0)g(y)d y −→ 0
Ω
m,n
(4.12)
m,n
as |x − x | → 0.
By Lemma 4.2, we deduce that
Tω(x) − Tω(x ) −→ 0,
as |x − x | −→ 0,
uniformly for all ω ∈ F. The result follows by Ascoli’s theorem.
(4.13)
738
Singular solutions for polyharmonic equation
Remark 4.4. Let α > 0. Then for 2m ≤ n and y ∈ Bc (0,2α) ∩ B, we have
lim
|x|→0
Gm,n (x, y)
= 0.
Gm,n (x,0)
(4.14)
So, using the same argument as in the proof of Lemma 4.3, we deduce that for
2m ≤ n,
Tω(x) −→ 0,
as |x| −→ 0,
(4.15)
uniformly for all ω ∈ F.
Proof of Theorem 4.1. We aim to show that there exists b0 > 0 such that for each
b ∈ (0,b0 ], there exists a continuous function u in B\{0} satisfying the following
integral equation:
u(x) = bGm,n (x,0) +
B
Gm,n (x, y) f y,u(y) d y,
x ∈ B \{0}.
(4.16)
Let β ∈ (0,1). Then by Lemma 4.3, the function
Tβ (x) =
1
Gm,n (x,0)
B
Gm,n (x, y)Gm,n (y,0)q y,βGm,n (y,0) d y
(4.17)
is continuous in B. Moreover, using (1.12), (H2 ), and (H3 ), we have
sup
B
ζ ∈B
δ(y)
δ(ζ)
m
Gm,n (ζ, y)g(y)d y g B .
(4.18)
So, we deduce by the dominated convergence theorem and (H2 ) that
lim Tβ (x) = 0 ∀x ∈ B.
(4.19)
β→0
Since the function β → Tβ (x) is nondecreasing in (0,1), it follows by Dini’s lemma
that
1
lim sup
β→0 x∈B Gm,n (x,0)
B
Gm,n (x, y)Gm,n (y,0)q y,βGm,n (y,0) d y = 0.
(4.20)
Imed Bachar et al. 739
Thus, there exists β ∈ (0,1) such that for each x ∈ B,
1
Gm,n (x,0)
1
Gm,n (x, y)Gm,n (y,0)q y,βGm,n (y,0) d y ≤ .
3
B
(4.21)
Let b0 = (2/3)β and b ∈ (0,b0 ]. We will use a fixed-point argument. Let
S = ω ∈ C(B) :
b
3b
≤ ω(x) ≤
.
2
2
(4.22)
Then, S is a nonempty, closed, bounded, and convex set in C(B). We define the
operator Γ on S by
Γω(x) = b +
1
Gm,n (x,0)
B
Gm,n (x, y) f y,ω(y)Gm,n (y,0) d y,
x ∈ B.
(4.23)
By Lemma 4.3, ΓS ⊂ C(B). Moreover, let ω ∈ S, then for any x ∈ B, we have
Γω(x) − b ≤ 3b
1
2 Gm,n (x,0)
b
≤ .
2
B
Gm,n (x, y)Gm,n (y,0)q y,βGm,n (y,0) d y
(4.24)
It follows that b/2 ≤ Γω(x) ≤ 3b/2 and so ΓS ⊂ S.
Next, we will prove the continuity of Γ in the uniform norm. Let (ωk )k be a
sequence in S which converges uniformly to ω ∈ S. Then since f is continuous
with respect to the second variable, we deduce by the dominated convergence
theorem that
Γωk (x) −→ Γω(x) as k −→ ∞, ∀x ∈ B.
(4.25)
Now, since ΓS is a relatively compact family in C(B), then
Γωk − Γω −→ 0
∞
as k −→ ∞.
(4.26)
So the Schauder fixed-point theorem implies the existence of ω ∈ S such that
Γω = ω.
For all x ∈ B, put u(x) = ω(x)Gm,n (x,0). Then, u is a continuous function in
B \{0} satisfying (4.16).
Furthermore, if 2m ≤ n, then by Remark 4.4, we obtain that lim|x|→0 ω(x) = b,
that is, lim|x|→0 u(x)/Gm,n (x,0) = b. This ends the proof.
Example 4.5. Let p > 0, λ < 2m, and µ < n ∧ 2m. Let V be a measurable function
in B such that for each x ∈ B,
V (x) ≤ δ(x)
λ
1
|x|µ
Gm,n (x,0)
p .
(4.27)
740
Singular solutions for polyharmonic equation
Then there exists b0 > 0 such that for each b ∈ (0,b0 ], the nonlinear problem
∆m u = (−1)m V (x)u p+1 (x)
u=
in B \ {0} (in the sense of distributions),
∂
∂m−1
u = · · · = m−1 u = 0
∂ν
∂ν
on ∂B,
(4.28)
has a positive solution u, continuous on B \{0} and satisfying for all x ∈ B,
3b
b
Gm,n (x,0) ≤ u(x) ≤ Gm,n (x,0)
2
2
(4.29)
and for 2m ≤ n, we have
lim
|x|→0
u(x)
= b.
Gm,n (x,0)
(4.30)
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Imed Bachar: Département de Mathématiques, Faculté des Sciences de Tunis, Campus
Universitaire, 1060 Tunis, Tunisia
E-mail address: Imed.Bachar@ipeit.rnu.tn
Habib Mâagli: Département de Mathématiques, Faculté des Sciences de Tunis, Campus
Universitaire, 1060 Tunis, Tunisia
E-mail address: habib.maagli@fst.rnu.tn
Syrine Masmoudi: Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire, 1060 Tunis, Tunisia
E-mail address: Syrine.Sassi@fst.rnu.tn
Malek Zribi: Département de Mathématiques, Faculté des Sciences de Tunis, Campus
Universitaire, 1060 Tunis, Tunisia
E-mail address: Malek.Zribi@insat.rnu.tn